Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3fffffff;
const long long mod = 1000000007;
const int maxn = 1e5 + 100;
int head[maxn * 2];
int sum[maxn];
struct edge {
  int to, next;
} e[maxn * 2];
int tol = 0;
void add(int u, int v) {
  e[++tol].to = v, e[tol].next = head[u], head[u] = tol;
}
vector<int> c[maxn];
int deep[maxn], fa[maxn][18];
int root[maxn];
struct node {
  int l, r, sum;
} T[maxn * 40];
int n, m;
namespace IO {
const int L = 1 << 24;
char buffer[L], *S, *T;
char outbuf[L], stk[40];
int os = 0;
inline char gc() {
  if (S == T) {
    T = (S = buffer) + fread(buffer, 1, L, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
inline bool read(int& x) {
  char c = gc();
  x = 0;
  for (; c < '-' && c != EOF; c = gc())
    ;
  if (c == EOF) return false;
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c - '0'), c = gc();
  return true;
}
inline void out() {
  fwrite(outbuf, 1, os, stdout);
  os = 0;
}
inline void outc(char c) {
  outbuf[os++] = c;
  if (os == L) out();
}
inline void outs(const char* s) {
  for (int i = 0; i < strlen(s); i++) outc(s[i]);
}
inline void outint(int x) {
  if (!x) outc('0');
  if (x < 0) outc('-'), x = -x;
  int i = 0;
  while (x) stk[++i] = x % 10 + '0', x /= 10;
  while (i) outc(stk[i--]);
}
}  // namespace IO
void bfs(int rt) {
  queue<int> q;
  deep[rt] = 0;
  fa[rt][0] = rt;
  q.push(rt);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = 1; i <= 17; i++) fa[t][i] = fa[fa[t][i - 1]][i - 1];
    for (int i = head[t]; i; i = e[i].next) {
      int v = e[i].to;
      if (v == fa[t][0]) continue;
      deep[v] = deep[t] + 1;
      fa[v][0] = t;
      q.push(v);
    }
  }
}
int lca(int u, int v) {
  if (deep[u] > deep[v]) swap(u, v);
  int hu = deep[u], hv = deep[v];
  int tu = u, tv = v;
  for (int det = hv - hu, i = 0; det; det >>= 1, i++)
    if (det & 1) tv = fa[tv][i];
  if (tu == tv) return tu;
  for (int i = 17; i >= 0; i--) {
    if (fa[tu][i] == fa[tv][i]) continue;
    tu = fa[tu][i];
    tv = fa[tv][i];
  }
  return fa[tu][0];
}
int cnt = 0;
void update(int l, int r, int& x, int y, int pos, int v) {
  T[++cnt] = T[y], T[cnt].sum += v, x = cnt;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(l, mid, T[x].l, T[y].l, pos, v);
  else
    update(mid + 1, r, T[x].r, T[y].r, pos, v);
}
void dfs(int u, int f) {
  sum[u] = sum[f] + (int)c[u].size();
  if (!c[u].empty()) {
    update(1, m, root[u], root[f], c[u][0], 1);
    for (int i = 1; i < c[u].size(); i++)
      update(1, m, root[u], root[u], c[u][i], 1);
  } else
    root[u] = root[f];
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    dfs(v, u);
  }
}
int query(int u, int v, int f, int k) {
  int l = 1, r = m;
  int lca_root = root[f];
  int x = root[u], y = root[v];
  while (l < r) {
    int mid = (l + r) / 2;
    int sum = T[T[x].l].sum + T[T[y].l].sum - 2 * T[T[lca_root].l].sum;
    int p1 = lower_bound(c[f].begin(), c[f].end(), l) - c[f].begin();
    int p2 = lower_bound(c[f].begin(), c[f].end(), mid + 1) - c[f].begin();
    sum += p2 - p1;
    if (k <= sum) {
      r = mid;
      x = T[x].l;
      y = T[y].l;
      lca_root = T[lca_root].l;
    } else {
      l = mid + 1;
      x = T[x].r;
      y = T[y].r;
      lca_root = T[lca_root].r;
      k -= sum;
    }
  }
  return l;
}
int main() {
  tol = 0;
  memset(head, 0, sizeof(head));
  int q;
  IO::read(n), IO::read(m), IO::read(q);
  for (int i = 1; i < n; i++) {
    int u, v;
    IO::read(u), IO::read(v);
    add(u, v);
    add(v, u);
  }
  for (int i = 1; i < m + 1; i++) {
    int v;
    IO::read(v);
    c[v].push_back(i);
  }
  for (int i = 1; i < m + 1; i++) sort(c[i].begin(), c[i].end());
  bfs(1);
  sum[0] = 0;
  dfs(1, 0);
  while (q--) {
    int u, v, a;
    IO::read(u), IO::read(v), IO::read(a);
    int f = lca(u, v);
    int sz = sum[u] + sum[v] - 2 * sum[f] + c[f].size();
    sz = min(sz, a);
    IO::outint(sz);
    IO::outc(' ');
    for (int i = 1; i < sz + 1; i++)
      IO::outint(query(u, v, f, i)), IO::outc(' ');
    IO::outc('\n');
  }
  IO::out();
  return 0;
}